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Suppose that $k$ is a field of characteristic $p$ such that $k$ is not a finite $k^p$-module. For example, $k = \mathbb{F}_p(x_1, x_2, x_3, ...)$.

Is it true that $k[[x]]$ is a free $(k[[x]])^p$-module? We know that it is flat by a theorem of Kunz. However, it is certainly not finite, so flat is not the same as free.

The naive thing to try (in terms of a basis) is to do the following. Choose {$\lambda_i$} a basis for $k$ over $k^p$. Consider the set {$\lambda_i x^j$} for $0 \leq j \leq p-1$. If $k$ is a finite $k^p$-vector space, this set is easily seen to be a basis for $k[[x]]$ over $(k[[x]])^p$.

However, because of our (ugly) field, we can consider the power series:

$$\lambda_0 + \lambda_1 x^1 + \lambda_2 x^2 + ... + \lambda_n x^n + ...$$

where say $\lambda_0, \lambda_1, ...$ runs over some countably infinite subset of the {$\lambda_i$}. It is easy to see that this cannot be written as a finite $(k[[x]])^p$-linear combination of subset of the $\lambda_i x^j$ (where again $0 \leq j \leq p-1$).

But of course, maybe there's some clever way to choose a basis that actually does work?

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  • $\begingroup$ $k^p=k\times\stackrel{p}\cdots\times k$? $\endgroup$
    – Fernando Muro
    Commented Apr 9, 2013 at 19:20
  • $\begingroup$ Dear @Fernando, $k^p$ means $\{\alpha^p:\alpha\in k\}$, i.e., the image of the Frobenius endomorphism of $k$. $\endgroup$
    – Keenan Kidwell
    Commented Apr 9, 2013 at 19:28
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    $\begingroup$ maybe not? since $\prod_{I} \mathbb{Z}$ is not free over $\mathbb{Z}$ when $I$ is not finite.. $\endgroup$
    – user20421
    Commented Apr 9, 2013 at 20:04
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    $\begingroup$ It seems to me that the larger ring $S$ is the $x$-adic completion of the extension by scalars along $k^p \to k$ of the power series ring $R$ over $k^p$. Since $k$ is a free $k^p$-module of infinite rank, the question becomes: is the $x$-adic completion of a free $R$-module of infinite rank also free? The answer to this should be no, as an infinite rank free $R$-module is never $x$-adically complete. $\endgroup$
    – anon
    Commented Apr 10, 2013 at 2:57
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    $\begingroup$ @Karl: By passing to direct summands (which preserves completeness), it suffices to show that $M :== \oplus_{i=1}^\infty R$ is not $x$-adically complete for any ring $R$ with a regular element $x$. Consider the elements $m_n:=\sum_{i=1}^n x^i \cdot e_i$, where $e_i$ is the $i$-th basis vector. Then $\{m_n\}$ is a Cauchy sequence in $M$ with no limit (as any $m := \sum_{i=1}^N a_i \cdot e_i \in M$ admits an open neighbourhood $m + x^{N+2} M$ that does not contain a single $m_n$). $\endgroup$
    – anon
    Commented Apr 10, 2013 at 15:04

1 Answer 1

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The answer to the question is no.

Let $R = k[[x]]$, $S = k[[y]]$, and $f:R \to S$ is the absolute Frobenius. We must show that $S$ is not a free $R$-module. By assumption on $k$, we know that $S$ must have infinite rank if it were free. On the other hand, $S$ is $x$-adically complete (since $x^p = y \in S$). Hence, the claim follows from:

Lemma: The $R$-module $M := R^{\oplus I}$ is $x$-adically complete if and only if $I$ is finite.

Proof: The "if" direction is clear. For the reverse, note that direct summands of complete $R$-modules are complete. By replacing $I$ with a subset, we may assume $I = \mathbf{N}$. Now consider the sequence $m_n := \sum_{i=1}^n x^i \cdot e_i \in M$, where $e_i$ is the $i$-th basis vector. This sequence is Cauchy (clear); I claim it has no limit in $M$. If there was a limit $m = \sum_{i=1}^N a_i \cdot e_i \in M$ with $a_i \in R$, then the neighbhourhood $m \in m + x^k M$ would contain infinitely many $m_n$ for any fixed $k$, but staring at the coefficient of $e_{N+1}$ shows this is false as long as $k \geq N+2$.

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  • $\begingroup$ @anon: could you specify your notationa bit? I guess you impose $y=x^p$ in the definition of $S$. But why is it $R$-adically complete, then? Thanks $\endgroup$
    – Filippo Alberto Edoardo
    Commented Apr 11, 2013 at 0:59
  • $\begingroup$ @Filippo: The $R$-submodule $x^p S \subset S$ is exactly the ideal $(x^p) \subset S$, so I don't understand the question. $\endgroup$
    – anon
    Commented Apr 11, 2013 at 5:18
  • $\begingroup$ ...neither do I, at least not anymore. I am sorry. $\endgroup$
    – Filippo Alberto Edoardo
    Commented Apr 25, 2013 at 17:36

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